What is the integral of t/((t^2+1)^{7/2)} ?

The integral of t/((t^2+1)^{7/2)} is -1/(5(t^2+1)^{5/2)}+C

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integral of t/((t^2+1)^{7/2)}

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{7}{2}}} d t

- \frac{1}{5 ( t ^{2} + 1 ) ^{\frac{5}{2}}} + C

Solution steps

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{7}{2}}} d t

Apply u-substitution

: \int \frac{1}{2 ( u + 1 ) ^{\frac{7}{2}}} d u

= \int \frac{1}{2 ( u + 1 ) ^{\frac{7}{2}}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{( u + 1 ) ^{\frac{7}{2}}} d u

Apply u-substitution

: \int \frac{1}{v ^{\frac{7}{2}}} d v

= \frac{1}{2} \cdot \int \frac{1}{v ^{\frac{7}{2}}} d v

Apply the Power Rule

: - \frac{2}{5 v ^{\frac{5}{2}}}

= \frac{1}{2} (- \frac{2}{5 v ^{\frac{5}{2}}})

Substitute back

= \frac{1}{2} (- \frac{2}{5 ( t ^{2} + 1 ) ^{\frac{5}{2}}})

Simplify

\frac{1}{2} (- \frac{2}{5 ( t ^{2} + 1 ) ^{\frac{5}{2}}}) : - \frac{1}{5 ( t ^{2} + 1 ) ^{\frac{5}{2}}}

= - \frac{1}{5 ( t ^{2} + 1 ) ^{\frac{5}{2}}}

Add a constant to the solution

= - \frac{1}{5 ( t ^{2} + 1 ) ^{\frac{5}{2}}} + C

y = 2 x^{2} - 1, (0, 3)

y^{^{''}} + 4 y^{^{'}} + 4 y = 4 e^{- 2 t}

\int \frac{( x - 3 )}{( x + 4 ) ( x - 5 )} d x

\frac{d}{d x} (x^{3} - 5 x^{2} + 7 x - 3)

\frac{\partial}{\partial x} (x^{2} y - \frac{1}{x y})

What is the integral of t/((t^2+1)^{7/2)} ?
The integral of t/((t^2+1)^{7/2)} is -1/(5(t^2+1)^{5/2)}+C